3.3.5 \(\int \text {csch}^6(c+d x) (a+b \sinh ^4(c+d x))^2 \, dx\) [205]

Optimal. Leaf size=84 \[ -\frac {b^2 x}{2}-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d} \]

[Out]

-1/2*b^2*x-a*(a+2*b)*coth(d*x+c)/d+2/3*a^2*coth(d*x+c)^3/d-1/5*a^2*coth(d*x+c)^5/d+1/2*b^2*cosh(d*x+c)*sinh(d*
x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3296, 1273, 1816, 213} \begin {gather*} -\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {b^2 \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {b^2 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4)^2,x]

[Out]

-1/2*(b^2*x) - (a*(a + 2*b)*Coth[c + d*x])/d + (2*a^2*Coth[c + d*x]^3)/(3*d) - (a^2*Coth[c + d*x]^5)/(5*d) + (
b^2*Cosh[c + d*x]*Sinh[c + d*x])/(2*d)

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 1273

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m
/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x],
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 3296

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*
x^2)^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a-2 a x^2+(a+b) x^4\right )^2}{x^6 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {\text {Subst}\left (\int \frac {-2 a^2+6 a^2 x^2-2 a (3 a+2 b) x^4+\left (2 a^2+4 a b+b^2\right ) x^6}{x^6 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {\text {Subst}\left (\int \left (-\frac {2 a^2}{x^6}+\frac {4 a^2}{x^4}-\frac {2 a (a+2 b)}{x^2}-\frac {b^2}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {b^2 x}{2}-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.60, size = 67, normalized size = 0.80 \begin {gather*} \frac {-4 a \coth (c+d x) \left (8 a+30 b-4 a \text {csch}^2(c+d x)+3 a \text {csch}^4(c+d x)\right )+15 b^2 (-2 (c+d x)+\sinh (2 (c+d x)))}{60 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4)^2,x]

[Out]

(-4*a*Coth[c + d*x]*(8*a + 30*b - 4*a*Csch[c + d*x]^2 + 3*a*Csch[c + d*x]^4) + 15*b^2*(-2*(c + d*x) + Sinh[2*(
c + d*x)]))/(60*d)

________________________________________________________________________________________

Maple [A]
time = 1.56, size = 140, normalized size = 1.67

method	result	size
risch	\(-\frac {b^{2} x}{2}+\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {4 a \left (15 b \,{\mathrm e}^{8 d x +8 c}-60 b \,{\mathrm e}^{6 d x +6 c}+40 a \,{\mathrm e}^{4 d x +4 c}+90 b \,{\mathrm e}^{4 d x +4 c}-20 a \,{\mathrm e}^{2 d x +2 c}-60 b \,{\mathrm e}^{2 d x +2 c}+4 a +15 b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4)^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*b^2*x+1/8/d*exp(2*d*x+2*c)*b^2-1/8/d*exp(-2*d*x-2*c)*b^2-4/15*a*(15*b*exp(8*d*x+8*c)-60*b*exp(6*d*x+6*c)+
40*a*exp(4*d*x+4*c)+90*b*exp(4*d*x+4*c)-20*a*exp(2*d*x+2*c)-60*b*exp(2*d*x+2*c)+4*a+15*b)/d/(exp(2*d*x+2*c)-1)
^5

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (76) = 152\).
time = 0.28, size = 267, normalized size = 3.18 \begin {gather*} -\frac {1}{8} \, b^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {16}{15} \, a^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {4 \, a b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4)^2,x, algorithm="maxima")

[Out]

-1/8*b^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 16/15*a^2*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c)
- 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 10*e^(-4*d*x -
4*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10
*c) - 1)) - 1/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10
*d*x - 10*c) - 1))) + 4*a*b/(d*(e^(-2*d*x - 2*c) - 1))

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (76) = 152\).
time = 0.38, size = 457, normalized size = 5.44 \begin {gather*} \frac {15 \, b^{2} \cosh \left (d x + c\right )^{7} + 105 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} - 4 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (105 \, b^{2} \cosh \left (d x + c\right )^{3} - {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 5 \, {\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 20 \, {\left (15 \, b^{2} d x - 2 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (63 \, b^{2} \cosh \left (d x + c\right )^{5} - 2 \, {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5 \, {\left (128 \, a^{2} + 96 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) - 20 \, {\left ({\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{4} + 30 \, b^{2} d x - 3 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 32 \, a^{2} - 120 \, a b\right )} \sinh \left (d x + c\right )}{120 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4)^2,x, algorithm="fricas")

[Out]

1/120*(15*b^2*cosh(d*x + c)^7 + 105*b^2*cosh(d*x + c)*sinh(d*x + c)^6 - (64*a^2 + 240*a*b + 75*b^2)*cosh(d*x +
c)^5 - 4*(15*b^2*d*x - 16*a^2 - 60*a*b)*sinh(d*x + c)^5 + 5*(105*b^2*cosh(d*x + c)^3 - (64*a^2 + 240*a*b + 75
*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 + 5*(64*a^2 + 144*a*b + 27*b^2)*cosh(d*x + c)^3 + 20*(15*b^2*d*x - 2*(15*
b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^2 - 16*a^2 - 60*a*b)*sinh(d*x + c)^3 + 5*(63*b^2*cosh(d*x + c)^5 - 2*
(64*a^2 + 240*a*b + 75*b^2)*cosh(d*x + c)^3 + 3*(64*a^2 + 144*a*b + 27*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 5
*(128*a^2 + 96*a*b + 15*b^2)*cosh(d*x + c) - 20*((15*b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^4 + 30*b^2*d*x -
3*(15*b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^2 - 32*a^2 - 120*a*b)*sinh(d*x + c))/(d*sinh(d*x + c)^5 + 5*(2
*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5*(d*cosh(d*x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c))

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**6*(a+b*sinh(d*x+c)**4)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (76) = 152\).
time = 0.49, size = 166, normalized size = 1.98 \begin {gather*} -\frac {60 \, {\left (d x + c\right )} b^{2} - 15 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 15 \, {\left (2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + \frac {32 \, {\left (15 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 40 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 20 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 60 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{2} + 15 \, a b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4)^2,x, algorithm="giac")

[Out]

-1/120*(60*(d*x + c)*b^2 - 15*b^2*e^(2*d*x + 2*c) - 15*(2*b^2*e^(2*d*x + 2*c) - b^2)*e^(-2*d*x - 2*c) + 32*(15
*a*b*e^(8*d*x + 8*c) - 60*a*b*e^(6*d*x + 6*c) + 40*a^2*e^(4*d*x + 4*c) + 90*a*b*e^(4*d*x + 4*c) - 20*a^2*e^(2*
d*x + 2*c) - 60*a*b*e^(2*d*x + 2*c) + 4*a^2 + 15*a*b)/(e^(2*d*x + 2*c) - 1)^5)/d

________________________________________________________________________________________

Mupad [B]
time = 0.75, size = 397, normalized size = 4.73 \begin {gather*} \frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (4\,a^2+3\,b\,a\right )}{5\,d}-\frac {4\,a\,b}{5\,d}-\frac {12\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {b^2\,x}{2}-\frac {\frac {4\,\left (4\,a^2+3\,b\,a\right )}{15\,d}-\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {\frac {8\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (4\,a^2+3\,b\,a\right )}{5\,d}+\frac {4\,a\,b}{5\,d}-\frac {16\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}-\frac {16\,a\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {8\,a\,b}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^4)^2/sinh(c + d*x)^6,x)

[Out]

(b^2*exp(2*c + 2*d*x))/(8*d) - ((4*exp(2*c + 2*d*x)*(3*a*b + 4*a^2))/(5*d) - (4*a*b)/(5*d) - (12*a*b*exp(4*c +
4*d*x))/(5*d) + (4*a*b*exp(6*c + 6*d*x))/(5*d))/(6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x)
+ exp(8*c + 8*d*x) + 1) - (b^2*x)/2 - ((4*(3*a*b + 4*a^2))/(15*d) - (8*a*b*exp(2*c + 2*d*x))/(5*d) + (4*a*b*e
xp(4*c + 4*d*x))/(5*d))/(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - (b^2*exp(- 2*c - 2*
d*x))/(8*d) - ((8*exp(4*c + 4*d*x)*(3*a*b + 4*a^2))/(5*d) + (4*a*b)/(5*d) - (16*a*b*exp(2*c + 2*d*x))/(5*d) -
(16*a*b*exp(6*c + 6*d*x))/(5*d) + (4*a*b*exp(8*c + 8*d*x))/(5*d))/(5*exp(2*c + 2*d*x) - 10*exp(4*c + 4*d*x) +
10*exp(6*c + 6*d*x) - 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) - 1) - (8*a*b)/(5*d*(exp(2*c + 2*d*x) - 1))

________________________________________________________________________________________